Calculate the amount of energy required to condense 0.500 kg of steam initially at 125*C to ice at -25*C.?

Question

I tried but I can't figure it out.

I know you do it in steps but I don't know what specific heat capacities to use

or even how to do this exactly

please help

Answer 1

There are five steps that need to be done to calculate the amount of energy that must be removed from this water to do this.  First, you need to cool the steam to the boiling point, 100C.  Next, you need to calculate the energy released as the steam condenses (heat of vaporization).  Next you need to cool the liquid water from the boiling point to the freezing point (0C).  Then you need to calculate the energy released as the water freezes (heat of fusion).  The final step is to cool the ice from the freezing point to the final temperature.

Step 1 (Steps 3 and 5 are the same, except for using the appropriate heat capacity, Cp, from the linked Wikipedia article):

Q = m * Cp * (Tf - Ti)

Q = 0.5kg * 2.08J/g-K * (100 - 125)K

Q = -2,600 J

Step 2:

Q = Hv * m / MW

Q = 40,650 J/mol * 0.5kg / 18.02

Q = -1,128,000 J

Step 4:

Q = Hf * m

Q = 333,550J/kg * 0.5kg

Q = -166,800 J

Hopefully that is enough direction to get you to the right answer.  Good luck!

Answer 2

The linked page lists a bunch of properties of water.

(Thanks to Uncle Al who answered a similar question about a year ago.)

Some average values from that page:

Specific heat capacity, ice: 2.108 kJ/kg-K

Specific heat capacity, water: 4.187 kJ/kg-K

Specific heat capacity, water vapor: 1.996 kJ/-kgK

Latent heat of melting - 334 kJ/kg

Latent heat of evaporation - 2,270 kJ/kg

Basically you:

-take the amount of energy to cool the steam 125 to 100C

-then the amount of energy to condense it to water

-then the amount of energy to cool the water 100C to  0C

-then the amount of energy to freeze it to ice

-then the amount of energy to cool the ice 0C to -25C

and add them all up.

Depending on your teacher/instructor, they may be happy if you assume values from the middle of the respective  temperature ranges you're using, or you can take the average value for the whole range. (If you had to justify the value you use vs just accepting the "average" values listed.)

I'll leave the math to you.